Unique determination for the extension of $f: A \to Y$ into the closure of A (Munkres 18.13) 18.13: "Let $A \subseteq X$. Let $F:A \to Y$ be continuous. Let $Y$ be Hausdorff. Prove that if $f$ is extended to a continuous function $g: \overline{A} \to Y$, then $g$ is uniquely determined by $f$."
Proof:
Let $g,h$ be extensions of $f$ to $\overline{A}$. 
Suppose that $A=\overline{A}$. Then $g,h=f$ trivially.
Now suppose that $A \subsetneq \overline{A}$. Then $A^{\prime}$ is nonempty. Let $x$ be a limit point of $A$. Suppose that $g(x)\neq h(x)$. Then, since $Y$ is Hausdorff, there exist neighborhoods $U,V$ of $g(x), h(x)$, respectively, such that $U \cap V = \emptyset$. 
By definition, $x \in g^{-1}(U) \cap h^{-1}(V)$. But then, since $g$ is continuous, this is a neighborhood of $x$, and since $x$ is a limit point, there exists $y\in g^{-1}(U) \cap h^{-1}(V) \cap A$. But then $g(y) \in U$ and $h(y) \in V$. However, since $y \in A$ as well, $f(y)=g(y)=h(y)$. This is a contradiction since $U \cap V = \emptyset$. Q.E.D
One part that I assumed is that if $x \in A$, $f(x)$ determines the value of any "extension." I'm worried that I didn't really use the 
Is this proof correct?
Note that this question has been asked before, I'm asking about my proof.
 A: The proof is essentially correct. You need not bother with limit points vs. points of the closure etc.: 
Suppose $g,h$ both continuously extend $f$ and are defined on $\overline{A}$, and suppose there is some $x \in \overline{A}$ such that $g(x) \neq h(x)$.
Using Hausdorffness of $Y$ we get $U$ and $V$ (disjoint, open, $g(x) \in U, h(x) \in V$ so $x \in g^{-1}[U] \cap h^{-1}[V]$, which is open by continuity of both $g$ and $h$, and we get $x'$ (better than $y$, as the latter suggests being in $Y$) in $A$ such that $x' \in A \cap g^{-1}[U] \cap h^{-1}[V]$ (because $x \in \overline{A}$ means every open neighbourhood of $x$ intersects $A$).
Then you use the fact that $g$ and $h$ extend $f$ when you say $f(x') = h(x') = g(x')$ as $x' \in A$, so $g$ and $h$ must agree with $f$ there. And then $f(x')$ contradicts disjointness of $U$ and $V$.
So all assumptions are used: namely Hausdorffness of $Y$, the fact that $g,h$ extend $f$, the fact that they are continuous and the fact that the domain is $\overline{A}$ (and not larger, because then we lose unicity). Of course such an extension of $f$ need not exist (consider $f(x) = \frac{1}{x}$ on $A = (0,1]$, where we cannot extend continuously to $0$, which lies in $\overline{A}$), but if it exists, it's uniquely determined by $f$ in the closure of $A$. 
Another proof if you know about nets (a generalisation of sequences): suppose $x \in \overline{A}$, then there is a net $(x_i)_{i \in I}$ from $A$ that converges to $x$, and then $g(x) = g(\lim_{i \in I} x_i) = \lim_{i \in I} g(x_i) = \lim_{i \in I} f(x_i) = \lim_{i \in I} h(x_i) = h(x)$, where continuity of $g$ and $h$ are used when we interchange limits and function calls, and $f(x_i) = h(x_i) = g(x_i)$ by extension. Then as limit of nets are unique in a Hausdorff space, $h(x) = g(x)$ and we are done. But this supposes that nets have been covered (which Munkres does not do, I believe) and we cannot use sequences in general spaces.  
A: Note that you never used that $x$ was a limit point, just a point in $\overline{A}\setminus A$. If $g$ and $h$ are not the same, then $h(x)\neq g(x)$ for some $x\in \overline{A}$. Necessarily $x\in\overline{A}\setminus{A}$ (why?). And then one can proceed exactly as you did.
